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Unformatted text preview: Muraj, Hamza Homework 27 Due: Apr 5 2006, 4:00 am Inst: Florin 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 3) 0 points A uniform rod, supported and pivoted at its midpoint, but initially at rest, has a mass of 84 g and a length 5 cm. A piece of clay with mass 34 g and velocity 1 . 7 m / s hits the very top of the rod, gets stuck and causes the clay- rod system to spin about the pivot point O at the center of the rod in a horizontal plane. Viewed from above the scheme is 1 . 7 m / s O 34 g 84 g (a) (b) (c) 5cm Figure: The piece of clay and rod: (a) before they collide, (b) at the time of the collision, and (c) after they collide. After the collisions the clay-rod system has an angular velocity about the pivot. With respect to the pivot point O , what is the magnitude of the initial angular momen- tum L i of the clay-rod system? Correct answer: 0 . 1445 kg m 2 / s. Explanation: Let : m putty = m = 34 g = 0 . 034 kg , m rod = 2 m = 84 g = 0 . 084 kg , = 5 cm = 0 . 05 m , and v = 1 . 7 m / s . Basic Concepts: Conservation of angular momentum ~ L ~ L = ~r ~p = m~r ~v X ~ ext = d ~ L dt L z = I . Therefore, if the net external torque acting on a system is zero, the total angular momentum of that system is constant. Since the total external torque acting on the clay-rod system is zero, the total angular momentum is a constant of motion. The total initial angular momentum L i is simply the angular momentum of the clay, since the rod is at rest initially L i = k ~r ~p k = m r v = m v 2 (1) = (0 . 034 kg)(1 . 7 m / s)(0 . 05 m) 2 = . 1445 kg m 2 / s . 002 (part 2 of 3) 0 points With respect to the pivot point O , what is the final moment of inertia I f of the clay-rod system? Correct answer: 8 . 75 10- 5 kg m 2 . Explanation: The final moment of inertia I f of the clay- rod system is the moment of inertia of the rod plus the moment of inertia of the clay I f = I rod + I clay = 1 12 2 m 2 + m 2 2 = 5 12 m 2 (2) = 5 12 (0 . 034 kg)(0 . 05 m) 2 = 8 . 75 10- 5 kg m 2 ....
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This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.
- Spring '09